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| JRevor (147) |  |
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http://www.cplusplus.com/reference/stl/set/ Reading in the reference. Compare: Comparison class: A class that takes two arguments of the same type as the container elements and returns a bool. My lecturer told a friend of mine, who told me set's Compare class could return an int +1 if a < b 0 if a = b -1 if a> b (or maybe it was the other way around, changing , -1 on the top, and 1 on the bottom) However, I can't find any help on this subject. That leads me to believe this information could be false. Any help? | |
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| R0mai (638) | |
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That's true to this point since int is convertible to bool. Reference is always right, professors not, and your friend might misunderstood something. Java (from languages which I know of) uses this int method thing in compareTo() method. | ||
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| firedraco (4744) | |
| For std::set's compare, it expects it to return a bool whether the first is > then second (or something similar). If you return an int, you could return it like that, but if you convert it to a bool, then you would get the wrong information, since 0 = false and non-zero = true, you would get true if they are not equal, and false if they are, creating a possibly weird sorting method. | |
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| JRevor (147) | |
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Thanks a lot for the help, ROmai and firedraco. still, i wonder how in the world is set going to know when two elements are equal, only having operator < overloaded . Does it compares attribute to attribute? What if the attributes are pointers to memory?(i have the intuition that i should overload operator = , in that case). | |
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| firedraco (4744) | |
| It doesn't care whether they are equal (since std::set can only have unique elements). | |
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| Abramus (170) | |
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!(a < b) && !(b < a) | ||
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| JRevor (147) | |
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That was awesome, Abramus! Thanks a lot, everyone. I'm marking this as solved. | |
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| kempofighter (1116) | |
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It doesn't! Set checks for equivalence, not equality and you need to understand the difference. For the case of a std::set<int> it makes no difference. In that case equivalence and equality will mean the same thing. For a set that contains user defined types the difference can be critical. That is a subject worth reading up on when you can make the time. | ||
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Confused about stl's set